Theorem 6.1: If a line is drawn parallel to one side of a triangle to intersect the other two side in distinct points, the other two sides are divided in the same ratio. Given: Ξ ABC where DE β₯ BC
To Prove: π΄π·/π·π΅ = π΄πΈ/πΈπΆ
Construction: Join BE and CD
Draw DM β₯ AC and EN β₯ AB.
Proof:
Now,
Now,
ar (ADE) = 1/2 Γ Base Γ Height
= 1/2 Γ AE Γ DM
ar (DEC) = 1/2 Γ Base Γ Height
= 1/2 Γ EC Γ DM
Divide (3) and (4)
"ar (ADE)" /"ar (DEC)" = (1/2 " Γ AE Γ DM" )/(1/2 " Γ EC Γ DM " )
"ar (ADE)" /"ar (DEC)" = "AE" /"EC"
Now,
βBDE and βDEC are on the same base DE
and between the same parallel lines BC and DE.
β΄ ar (BDE) = ar (DEC)
Hence,
"ar (ADE)" /"ar (BDE)" = "ar (ADE)" /"ar (DEC)"
"AD" /"DB" = "AE" /"EC"
Hence Proved.

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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